Question

$\lim _{x \rightarrow 1} \frac{n x^{n+1}-(n+1) x^{n}+1}{\left(e^{x}-e\right) \sin \pi x}$ $\lim _{h \rightarrow 0} \frac{n(1+h)^{n+1}-(n+1)(1+h)^{n}+1}{-e\left(e^{h}-1\right) \sin \pi h}$ $\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{n}(1+\mathrm{h})^{\mathrm{n}+1}-(\mathrm{n}+1)(1+\mathrm{h})^{\mathrm{n}}+1}{-\pi \mathrm{t} \mathrm{h}^{2}}$ $\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{n}(\mathrm{n}+1)(1+\mathrm{h})^{\mathrm{n}}-\mathrm{n}(\mathrm{n}+1)(1+\mathrm{h})^{\mathrm{n}-1}}{-2 \pi \mathrm{t} h}$ $\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n})(1+\mathrm{h})^{\mathrm{n}-1}-\mathrm{n}(\mathrm{n}+1)(1+\mathrm{h})^{\mathrm{n}-2}(\mathrm{n}-1)}{-2 \pi \mathrm{e}}$ $=\frac{\mathrm{n}^{2}(\mathrm{n}+1)-\mathrm{n}(\mathrm{n}-1)}{-2 \pi \mathrm{e}}(\mathrm{n}+1)=\frac{\mathrm{n}(\mathrm{n}+1)}{-2 \pi \mathrm{e}}$ Put \(\mathrm{n}=100\) \(=\frac{-5050}{\pi \mathrm{e}}\)

Short answer

Answer: The limit is \(\frac{-5050}{\pi e}\).

Step-by-step solution

Choose the correct limit expression

We are given four equivalent forms of a limit expression. All four forms have the same limit. We can use the last given form to solve the problem: \(\lim _{h \rightarrow 0} \frac{n(n+1)(n)(1+h)^{n-1}-n(n+1)(1+h)^{n-2}(n-1)}{-2\pi e}\)

Substitute n with the given value

We need to compute the limit when n = 100. So we substitute n with 100 in the expression: \(\lim _{h \rightarrow 0} \frac{100(100+1)(100)(1+h)^{100-1}-100(100+1)(1+h)^{100-2}(100-1)}{-2\pi e}\)

Compute the limit

As h approaches 0, this expression simplifies as follows: \begin{align*} \lim _{h \rightarrow 0} \frac{100(101)(100)(1+h)^{99}-100(101)(1+h)^{98}(99)}{-2\pi e} &= \frac{100(101)(100)(1)^{99}-100(101)(1)^{98}(99)}{-2\pi e} \\ &= \frac{100\cdot101\cdot100-100\cdot101\cdot99}{-2\pi e} \\ &= \frac{100\cdot101}{-2\pi e} \end{align*} This is our final answer:

Final Answer

The limit is equal to \(\frac{-5050}{\pi e}\).